Problem: You have found the following ages (in years) of all 6 turtles at your local zoo: $ 21,\enspace 84,\enspace 21,\enspace 73,\enspace 60,\enspace 43$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{21 + 84 + 21 + 73 + 60 + 43}{{6}} = {50.3\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $21$ years $-29.3$ years $858.49$ years $^2$ $84$ years $33.7$ years $1135.69$ years $^2$ $21$ years $-29.3$ years $858.49$ years $^2$ $73$ years $22.7$ years $515.29$ years $^2$ $60$ years $9.7$ years $94.09$ years $^2$ $43$ years $-7.3$ years $53.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{858.49} + {1135.69} + {858.49} + {515.29} + {94.09} + {53.29}} {{6}} $ $ {\sigma^2} = \dfrac{{3515.34}}{{6}} = {585.89\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{585.89\text{ years}^2}} = {24.2\text{ years}} $ The average turtle at the zoo is 50.3 years old. There is a standard deviation of 24.2 years.